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Let $F$ be a field of characteristic 2. I need to show the existence of a quadratic polynomial in $F[t]$ which cannot be solved by adjoining all square roots of elements in the field.

Attempt:

For $F=\mathbb{Z}_2$, $f(t)=t^2+t+1$ works since $\mathbb{Z}_2$ is closed taking square roots.

I don't know how to do the general case. I think that the same polynomial could work.

Edit: Robert shows a counterexample in comments.

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  • $\begingroup$ In a field $\mathbb F$ of characteristic $2$, every element has a square root, and so adjoining the square roots of all the elements of $\mathbb F$ to $\mathbb F$ does not help in the least. What you need to do is find a quadratic polynomial that is irreducible over $\mathbb F$. $\endgroup$ – Dilip Sarwate Jul 25 '13 at 0:36
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    $\begingroup$ @DilipSarwate $X$ does not have a square root in ${\bf F}_2(X)$. The point is that forming $L/F$ by adjoining square roots should be expected to hurt our chances of finding a quadratic in $F[t]$ irreducible over $L$, in fact, in characteristic $\ne2$ it's fatal since then the quadratic formula applies, and the exercise asks us to overcome these apparent grim chances. $\endgroup$ – anon Jul 25 '13 at 0:40
  • $\begingroup$ @anon What is $T$? That is, what properties do we ascribe to $T$ other than it is something that we can add to or multiply by $0$ and $1$? And how can we be sure that $\mathbb F_2(T)$ is a field? For example, $\mathbb F_2[x]$, the set of all polynomials in an indeterminate $x$ with coefficients in $\mathbb F_2$ is a ring and not a field. $\endgroup$ – Dilip Sarwate Jul 25 '13 at 0:47
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    $\begingroup$ @DilipSarwate I am using standard notation to refer to the field of rational functions in the variable $X$ (I originally had the letter $T$ in my comment) with coefficients from ${\bf F}_2$, the field with just $0$ and $1$. The relevant property we ascribe to $X$ (or $T$, or whatever capital letter we want to use) is that it is transcendental over ${\bf F}_2$. Showing the ring of rational functions with coefficients from a field is itself a field is a basic exercise for students. I can't say I understand why you're asking me about these things. $\endgroup$ – anon Jul 25 '13 at 0:51
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    $\begingroup$ The polynomial $f(t)=t^2+t+1$ doesn't work if we consider $F$ to be its splitting field. Also, do there need to be more hypotheses? If $F$ is algebraically closed then there exists no such $f$. $\endgroup$ – rfauffar Jul 25 '13 at 1:00
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If $F$ is algebraically closed, then there exists no such $f$, since any $f$ splits. The polynomial $f(t)=t^2+t+1$ works for $\mathbb{Z}_2$, but not in general (take $F$ to be the splitting field of this polynomial). Maybe there are other conditions you can impose on $F$ so that this is true?

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    $\begingroup$ $F$ finite is an enough extra-condition. If $F$ is finite all elements have square roots in $F$ since $t\mapsto t^2$ is onto. But $t\mapsto t+t^2$ is not onto, then there is some $a\in F$ such that $t^2+t+a$ has no roots. $\endgroup$ – Gaston Burrull Jul 30 '13 at 6:01
  • $\begingroup$ Nice! That's enough then... $\endgroup$ – rfauffar Jul 30 '13 at 12:28

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